Is there a simple way to show that the polynomial $P(x) = x^8+6x^4+1$ is irreducible over $\mathbb Q$ by hand? The CAS software Sage tells me that it indeed is. It appears as polynomial 8.0.4194304.1 in the LMFDB database.
I tried to translate the variable and apply Eisenstein criterion without success. Cohn's criterion shows that $P$ is irreducible if $100060001$ is prime, which it is, but it seems tedious to show that $100060001$ is prime by hand.
Here is an approach using sufficient and necessary condition for irreducibility of $f(x^n)$:
Theorem. Let $f(x)\in\mathbb{Q}[x]$ be irreducible polynomial and let $\alpha\in \mathbb{C}$ be any of its roots. Then $f(x^n)$ is irreducible in $\mathbb{Q}[x]$ if and only if $\alpha\not\in \mathbb{Q}(\alpha)^p$ for all primes $p\mid n$ and $\alpha \not\in -4 \mathbb{Q}(\alpha)^4$ when $4\mid n$.
Proof. Combination of Capelli Lemma and the Vahlen-Capelli irreducibility criterion.
In our case let $f(x)=x^2+6x+1$ and we want to check irreducibility of $f(x^4)$. Clearly $f(x)$ is irreducible as it is quadratic and has no rational roots, the roots are $\alpha=-3+ 2\sqrt{2}$ and $\beta=-3- 2\sqrt{2}$. By the theorem above, we need to check two things:
In this case also notice $\mathbb{Q}(\alpha)=\mathbb{Q}(\sqrt{2})$.
Ad 1) For the sake of contradiction, assume $\alpha=-3+ 2\sqrt{2}=(p\sqrt{2}+q)^2$ for some $p,q\in \mathbb{Q}$. Expanding and comparing the coefficients we get in particular $-3=2p^2+q^2$, which is clearly impossible - the right side is non-negative.
Ad 2) Similarly assume $\alpha=-4(p\sqrt{2}+q)^4$, expanding and comparing we get in particular $-3=-16p^4-48p^2q^2-4q^4$. By completing the square (with a little twist in the sign), we can rewrite this as $$3=(4p^2-2q^2)^2+(8pq)^2.$$ This is impossible as $3$ cannot be written as a sum of rational squares (multiply both sides by common denominator to get a Diophantine equation in a form $3u^2=v^2+w^2$, for more details see for example Show that the curve $x^2+y^2-3=0$ has no rational points ).
Therefore by the above theorem, $f(x^4)=x^8+6x^4+1$ is irreducible over rationals. In fact, the theorem implies that $x^{2^{m+1}}+6x^{2^m}+1$ is irreducible for all $m\geq 0$.
See also:
I have no idea what would be the simplest argument. I'm in the habit of milking everything I can out of a few modular considerations and symmetries, so let me proffer the following.
Assume contrariwise that $P(x)$ factors non-trivially in $\Bbb{Q}[x]$ By Gauss's lemma and friends we deduce that it then also factors in $\Bbb{Z}[x]$, and hence by reduction modulo $3$ also in $\Bbb{F}_3[x]$. But modulo three we have $$P(x)\equiv x^8+1.$$ So if $\alpha$ is a root of $P(x)$ in some extension field $K$ of $\Bbb{F}_3$, we have $\alpha^8=-1$, and then $\alpha^{16}=1$. As two is the only prime factor of sixteen, together these imply that $\alpha$ is a root of unity of order exactly sixteen. Applying Lagrange's theorem to the multiplicative group $K^*$ we deduce that $16\mid |K|-1$. As $16\mid 3^4-1$ but $16\nmid 3^2-1$, we can conclude that $[K:\Bbb{F}_3]=4$. In other words, the irreducible factors of $P(x)$ in $\Bbb{F}_3[x]$ must have degree four.
The next trick is the Sophie Germain factorization $$a^4+4=(a^4+4a^2+4)-4a^2=(a^2+2a+2)(a^2-2a+2).$$ We have $4\equiv1\pmod3$, so this gives ($a=x^2$) $$P(x)\equiv x^8+4=(x^4+2x^2+2)(x^4+2x^2-2)=(x^4-x^2-1)(x^4+x^2-1).$$ In view of the preceding paragraph these quartic factors are irreducible in $\Bbb{F}_3[x]$. At this point we can already conclude that the only possible factorization of $P(x)$ is into a product of two quartics, $G(x)$ and $H(x)$, such that $$G(x)\equiv x^4-x^2-1\pmod3\qquad\text{and}\qquad H(x)\equiv x^4+x^2-1\pmod3.$$ Looking at the constant term of $P(x)=G(x)H(x)$ we conclude that the constant terms of both $G$ and $H$ must be equal to $-1$. So we can write $$G(x)=x^4+Ax^3+Bx^2+Cx-1$$ with $A,B,C\in\Bbb{Z}.$
The first obvious symmetry $P(x)$ has is that it is palindromic (=stays the same, if you write the sequence of coefficients backwards), in other words $P(x)=x^8P(\dfrac1x)$. If $z$ is a complex root $G(x)$, then $P(z)=0$ and by the palindromic property also $P(1/z)=0$. Meaning that $1/z$ has to be a zero of either $G(x)$ or $H(x)$. But $$0=z^{-4}G(z)=1+A(1/z)+B(1/z)^2+C(1/z)^3-(1/z)^4,$$ so the minimal polynomial of $1/z$ must be $R(x)=x^4-Cx^3-Bx^2-Ax-1$ (rescaled to be monic). If $R(x)=G(x)$ then $B=0$. This is impossible because earlier we figured out that $B\equiv-1\pmod3$. The remaining possibility is that $R(x)=H(x)$, in other words $$H(x)=x^4-Cx^3-Bx^2-Ax-1.$$
A second symmetry we spot right away is that $P(x)$ only contains even degree terms. This implies that $P(-z)=P(z)=0$. Let's play the same game again. The number $-z$ is a root of $$G(-x)=x^4-Ax^3+Bx^2-Cx-1,$$ and because $-z$ is a root of $P(x)$ we can conclude that either I) $G(-x)=G(x)$ or II) $G(-x)=H(x)$.
I) The identity $G(-x)=G(x)$ implies that $A=C=0$, so $G(x)=x^4+Bx^2-1$ and $H(x)=x^4-Bx^2-1$, when $$ G(x)H(x)=(x^4-1)^2-(Bx^2)^2=x^8-(2+B^2)x^4+1. $$ Because $6=-2-B^2$ for all $B\in\Bbb{Z}$, this is impossible, so factorizations of type I) do not exist.
II) The identity $G(-x)=H(x)$ implies $B=0$ (and also $A=C$), but we earlier figured out that $B\equiv-1\pmod3$, so this is also impossible.
The conclusion is that $P(x)$ is irreducible.
Over $\Bbb R$ your polynomial can be reduced in the following way:
$\begin{align}&x^8+6x^4+1=\\[3pt]&\!=\!x^8+6x^4+9-8=\left(x^4+3\right)^2-\big(2\sqrt2\big)^2=\\[3pt]&\!=\!\left(x^4+3+2\sqrt2\right)\left(x^4+3-2\sqrt2\right)=\\[3pt]&\!=\!\bigg[(x^2)^2\!+(\sqrt2\!+\!1)^2\!+2x^2\!\left(\sqrt2\!+\!1\right)-2x^2\!\left(\sqrt2\!+\!1\right)\bigg]\\[3pt]&\quad\bigg[(x^2)^2\!+(\sqrt2\!-\!1)^2\!+2x^2\!\left(\sqrt2\!-\!1\right)-2x^2\!\left(\sqrt2\!-\!1\right)\bigg]=\\[3pt] &\!=\!\bigg[\left(x^2\!+\sqrt2+1\right)^2\!-\left(x\sqrt{2\sqrt2+2}\right)^2\bigg]\\[3pt]&\quad\bigg[\left(x^2\!+\sqrt2-1\right)^2\!-\left(x\sqrt{2\sqrt2-2}\right)^2\bigg]=\\[3pt]&\!=\!\left(x^2\!\!+\!x\sqrt{2\sqrt2\!+\!2}+\!\sqrt2\!+\!1\right)\!\left(x^2\!\!-\!x\sqrt{2\sqrt2\!+\!2}+\!\sqrt2\!+\!1\right)\\[3pt]&\left(x^2\!\!+\!x\sqrt{2\sqrt2\!-\!2}\!+\!\sqrt2\!-\!1\right)\!\left(x^2\!\!-\!x\sqrt{2\sqrt2\!-\!2}\!+\!\sqrt2\!-\!1\right).\end{align}$
Any possible factorization of $x^8+6^4+1$ over $\Bbb Q$ should contain a factor which can be obtained by multiplying some factors of the previous factorization over $\Bbb R$, but in no way we can get a factor over $\Bbb Q$ (whose degree is less than $8$) by multiplying the previous factors over $\Bbb R$, hence the polynomial $x^8+6x^4+1$ is irreducible over $\Bbb Q$.
As Jyrki Lahtonen pointed out, the polynomial is invariant under the transformation $x^{deg(f)}f(1/x)$ and $f(-x)=f(x)$. Hence, it can be shown that the polynomial can be obtained by rational transformation of the polynomial $x^2+4$ (which is irreducible) and $Q(x)=\frac{x^4+1}{x^2}$, that is, $$x^8+6x^4+1=(x^2)^2\cdot ((\frac{x^4+1}{x^2})^2+4).$$
In a recent arxiv paper (https://arxiv.org/abs/2306.13502) it is shown that the factorization of such rational transformations is very regular in the sense that its factors are given by a single $G$-orbit, where $G=\{x,-x,1/x,-1/x\}$ under composition and the action is the rational transformation of polynomials with elements out of $G$.
Therefore, the only possible factorizations over $\mathbb{Q}[x]$ that $x^8+6x^4+1$ can have is one of the following:
And it can be done "by hand" that 2.,3. and 4. cannot happen.
$x^2+6x+1$ has negative real roots, so $P(x)=x^8 + 6 x^4+1$ has no real roots, that is, it is positive on $\mathbb{R}$. Assume it were reducible over $\mathbb{Z}$; then $P(x) =P_1(x) P_2(x)$, $P_1(x)$, $P_2(x)$ monic, with integral coefficients, positive on $\mathbb{R}$, But look at the values of $P(x)$ at $x=2,4,6,8,10$ ( so also $-2$, $\ldots$, a total of $10$ values, they are all primes ( not done by hand). Moreover the value at $0$ is $1$. In fact, it would have been enough to identify $7=8-1$ points where $P(x)$ takes a prime value (although they do come in opposite pairs). That means for at least one of the factors $P_i(x)$ ( of degree $d_i$) the number of solutions of $P_i(x) = 1$ is at least $d_i$. Combined with guaranteed value $1$ at $0$ this gives $d_i+1$ points where $P_i(x)$ takes value $1$, not possible ( would make $P_1(x)$ a constant).
The polynomial $f(x) = x^2 + 6x + 1$ can be factored as $f(x) = (x + 3 - 2\sqrt2)(x + 3 + 2 \sqrt 2)$. Therefore any multiplicative combination of the irreducible polynomials of $P(x) = f(x^4)$ over $\mathbb C$ will contain an irrational number, except if they are all multiplied together, hence the irreducibility over $\mathbb Q$.
